Problem: Rewrite the equation by completing the square. $x^{2} +3 x -28 = 0$ $(x + $
Explanation: $\begin{aligned} x^2 +3 x -28&=0 \\\\ x^2 +3 x&=28 \end{aligned}$ Now we want to complete $x^2 +3 x$ into a perfect square. To do that, we should add $\left(\dfrac{{3}}{2}\right)^2={\dfrac{9}{4}}$ to it: $x^2{+3}x + {\dfrac{9}{4}}=\left(x +\dfrac{3}{2} \right)^2$ $\begin{aligned} x^2 +3 x&=28 \\\\ x^2 +3 x + {\dfrac{9}{4}}&=28 + {\dfrac{9}{4}} \\\\ \left(x +\dfrac{3}{2} \right)^2&=\dfrac{121}{4} \end{aligned}$ In conclusion, the equation after completing the square is written as: $\left(x +\dfrac{3}{2} \right)^2=\dfrac{121}{4}$